A bag contains $3$ red jelly beans, $3$ green jelly beans, and $6$ blue jelly beans. If we choose a jelly bean, then another jelly bean without putting the first one back in the bag, what is the probability that the first jelly bean will be blue and the second will be green? Write your answer as a simplified fraction.
The probability of event A happening, then event B, is the probability of event A happening times the probability of event B happening given that event A already happened In this case, event A is picking a blue jelly bean and leaving it out. Event B is picking a green jelly bean. Let's take the events one at at time. What is the probability that the first jelly bean chosen will be blue? There are $6$ blue jelly beans, and $12$ total, so the probability we will pick a blue jelly bean is $\dfrac{6} {12}$ After we take out the first jelly bean, we don't put it back in, so there are only $11$ jelly beans left. Since the first jelly bean was blue, there are still $3$ green jelly beans left. So, the probability of picking a green jelly bean after taking out a blue jelly bean is $\dfrac{3} {11}$ Therefore, the probability of picking a blue jelly bean, then a green jelly bean is $\dfrac{6}{12} \cdot \dfrac{3}{11} = \dfrac{3}{22}$